Question

Assume that you have 1 mL of a solution of amylase (an enzyme) at a concentration of 15 mg protein/mL. Calculate the volume of diluting buffer that you would have to add to 1.0 mL of the amylase stock solution if you wished the final concentration of the solution to be 345 µg protein/mL.

Answer 1

Answer:

42,5 mL

Explanation:

We need to use the serial dilution formula beacuse we start with a stock concentrate solution and we need to prepare a new less concentrated one.

$DF=\frac{Vi}{Vf}$

DF in the dilution factor, Vi is the initial volume and Vf is the final volume.

The first step is to have the same measurment unit so we need to convert 345 µg to mg.

we know that 1 µg equals 0,001 g, hence:

$345 µg = 0,345 mg$

now the final volume is 0,345 mg  protein/ mL and the inital volume is 15mg protein/mL, both of them are in the same unit so we can use the formula

$DF= \frac{15mg protein/mL}{0.345mg protein/mL}$

$DF= 43,5 mg protein/ mL$

Now since the question said that we already have 1.0mL of the amylase stock solution we need to subtract that 1.0mL to the 43,5 mg protein/mL

$43,5mL-1,0mL = 42,5 mL$

So, we need 42,5 mL of diluting buffer if we want a final concentration of 345 µg protein/mL (0.345 mg protein/mL)